Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 4}{x + 8} = \dfrac{-5x + 28}{x + 8}$
Explanation: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 4}{x + 8} (x + 8) = \dfrac{-5x + 28}{x + 8} (x + 8)$ $ x^2 + 4 = -5x + 28$ Subtract $-5x + 28$ from both sides: $ x^2 + 4 - (-5x + 28) = -5x + 28 - (-5x + 28)$ $ x^2 + 4 + 5x - 28 = 0$ $ x^2 - 24 + 5x = 0$ Factor the expression: $ (x - 3)(x + 8) = 0$ Therefore $x = 3$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.